Given a cyclic $ABCD$ with $AB=AD$. Points $M$ and $N$ are marked on the sides $CD$ and $BC$, respectively, so that $DM+BN=MN$. Prove that the circumcenter of the triangle $AMN$ belongs to the segment $AC$.
N.Sedrakian
Take point $H\in MD^\rightarrow$ such that $DH=BN$. $$AB=AD\ \wedge \ \angle ABN=180^\circ-\angle ADC=\angle{ADH}\ \wedge\ BN=DH\iff\triangle ABN\equiv ADH\implies AN=AH$$$$AN=AH\ \wedge\ AM=AM \wedge\ NM=BN+DM=DH+DM=MH\implies \triangle ANM\equiv\triangle AHM$$Hence rays $NA^\rightarrow,MA^\rightarrow$ bisect respectively angles $BNM,NMD$. Let $I$ be the incenter of triangle $NCM$. Then $$NI\perp\ AN\wedge\ MI\perp AM$$Since $AB=AD$ ray $CA^\rightarrow$ bisects angle $NCM$ hence we know point $I$ belongs to the segment $AC$. Consider homothethy centered at $A$ with scale $1/2$. Image of $I$ is the circumcenter of the triangle $AMN$. QED